Results 1 to 7 of 7

Math Help - Convergence

  1. #1
    Senior Member
    Joined
    Apr 2009
    Posts
    306

    Convergence

    Does \displaystyle{\prod_{\text{primes p}}\left(1+\frac{1}{p}\right)} converge, and if so to what value?

    I'm not too sure how to go about this question, any help would be greatly appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    The Swiss mathematician Leonhard Euler demonstrated two and half centuries ago that the series...

    \displaystyle \sum_{p} \frac{1}{p}

    ... diverges, so that diverges also the product...

    \displaystyle \prod_{p} (1+\frac{1}{p})

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    However, if s>1 then,

    \displaystyle\prod_{p}\left(1+\dfrac{1}{p^s}\right  )=\dfrac{\zeta(s)}{\zeta{(2s)}}

    So, can extend the formula,

    \displaystyle\prod_{p}\left(1+\dfrac{1}{p}\right)=  \dfrac{\zeta(1)}{\zeta{(2)}}=\dfrac{+\infty}{\pi^2/6}=+\infty
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Apr 2009
    Posts
    306
    @chisigma thanks for the help! however I am still a bit unsure on how the divergence of the sum implies that the product also diverges?

    @FernandoRevilla cheers for that! I haven't seen that formula yet, nor can I find it in my notes, however im very interested, can you perhaps show me a proof of how the formula comes about?

    thanks again!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    Quote Originally Posted by usagi_killer View Post
    @FernandoRevilla cheers for that! I haven't seen that formula yet, nor can I find it in my notes, however im very interested, can you perhaps show me a proof of how the formula comes about?

    We have for every s>1:

    \dfrac{\zeta (s)}{\zeta (2s)}=\dfrac{\prod_{p} (1-p^{-s})^{-1}}{\prod_{p} (1-p^{-2s})^{-1}}=\ldots
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Quote Originally Posted by usagi_killer View Post
    @chisigma thanks for the help! however I am still a bit unsure on how the divergence of the sum implies that the product also diverges?
    In...

    Convergence of Infinite Products The Everything Seminar

    ... there is the proof of the following theorem: an 'infinite product' of the form...

    \displaystyle  \prod_{n=1}^{\infty} (1+a_{n})

    ... converges if and only if the 'infinite sum'...

    \displaystyle  \sum_{n=1}^{\infty} a_{n}

    ... converges
    ...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Apr 2009
    Posts
    306
    Thanks guys! Makes sense now!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: May 1st 2010, 09:22 PM
  2. dominated convergence theorem for convergence in measure
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: December 5th 2009, 04:06 AM
  3. Replies: 6
    Last Post: October 1st 2009, 09:10 AM
  4. series convergence and radius of convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 15th 2008, 08:07 AM
  5. Replies: 6
    Last Post: October 24th 2008, 01:45 PM

Search Tags


/mathhelpforum @mathhelpforum